SLAC: Statistical total lesion metabolic activity computation by fuzzy unsupervised learning of PET images

Abstract

Accurate lesion metabolic response estimation is imperative for efficient tumor staging and follow-up studies. Positron emission tomography (PET) successfully images the lesion metabolic activity. Nonetheless, on course of accurate delineation, chances are high to end up with activity underestimation as, due to the limited resolution, the PET images suffer from partial volume effects. Recently, PET images were modeled as a fuzzy mixture to delineate lesions accurately. We extend this work by proposing a statistical lesion activity computation (SLAC) approach to robustly estimate the total lesion activity (TLA) directly from the modeled partial volume mixtures, without an explicit delineation. To evaluate the proposed method, PET scans of phantoms containing spherical and non-spherical lesions with increased activity uptake were simulated. The PET images were reconstructed with the standard clinically used maximum likelihood expectation maximization and an edge preserving maximum a posteriori (MAP) algorithm, both with resolution recovery. From these images, the TLA was estimated in each lesion using the proposed method and compared to the TLA estimation in the tumor delineations obtained with three state-of-the-art PET delineation schemes. SLAC outperformed TLA estimation via tumor delineation and showed robust against variation in reconstruction parameters. With reference to the ground truth knowledge, SLAC gives median $\delta$TLA$~\approx$ 5 % for spherical lesions. For more realistic non-spherical lesions, median $\delta$TLA$~\approx$ 15 %.

Appendices

Appendix A: Parameter estimation

1.1 Gaussian distribution

Let $\mathcal Y$ be Gaussian distributed $\mathcal N (\mu ,\sigma ^2)$ with the probability density function $f_\mathcal{Y }(x) \!=\! \frac{1}{\sqrt{2\pi \sigma ^2}} \exp \left(\!-\frac{(x-\mu )^2}{2\sigma ^2}\!\right)$ and the characteristic function $\varphi _\mathcal{Y }(\omega ) \!=\! \exp (j\mu \omega -\frac{1}{2}\sigma ^2\omega ^2)$, then from a $\mathcal N (\mu ,\sigma ^2)$ population with samples $\mathbf x \!=\!\left\{ x_s \right\} _{s \in \mathcal S }$ where $\mathcal S \! =\! \{1,\ldots ,N\}$, the maximum likelihood estimates (MLE) are given by

$$\begin{aligned} \widehat{\mu } = \frac{1}{N}\sum _{i=1}^{N} x_i, \quad \widehat{\sigma ^2} =\frac{1}{N}\sum _{i=1}^{N} (x_i-\widehat{\mu })^2 \end{aligned}$$

(22)

1.2 Laplace distribution

Let $\mathcal Y$ be Laplace distributed $\mathcal L (\mu ,\sigma )$ with the probability density function $f_\mathcal{Y }(x)=\frac{1}{2\sigma } \exp \left(-\frac{|x-\mu |}{\sigma }\right)$ and the characteristic function $\varphi _\mathcal{Y }(\omega )= \frac{\exp (j\mu \omega )}{1+\sigma ^2\omega ^2}$, then from a $\mathcal L (\mu ,\sigma )$ population with samples $\mathbf x \!=\!\left\{ x_s \right\} _{s \in \mathcal S }$ where $\mathcal S \! =\! \{1,\ldots ,N\}$, the maximum likelihood estimates (MLE) are given by

$$\begin{aligned} \widehat{\mu } = \text{ Median} (\mathbf x ), \quad \widehat{\sigma } =\frac{1}{N}\sum _{i=1}^{N} \left|{x}_i- \widehat{\mu } \right| \end{aligned}$$

(23)

Appendix B: Convex combination of random variables

1.1 Derivation of probability density function

Let $\mathcal Y _0$ and $\mathcal Y _1$ be two random variables with probability density function $f_\mathcal{Y _0}(x)$ and $f_\mathcal{Y _1}(y)$, respectively, then their convex combination $\mathcal Y _s=\alpha \mathcal Y _0+\beta \mathcal Y _1$, will be distributed with density $f_\mathcal{Y _s}(z)$ as,

$$\begin{aligned} f_\mathcal{Y _s}(z)&= \frac{\mathrm{d}}{\mathrm{d}z} \int \int _{\alpha x + \beta y\le z} f_\mathcal{Y _0\mathcal Y _1}(x,y)\mathrm{d}x\mathrm{d}y \nonumber \\&= \frac{\mathrm{d}}{\mathrm{d}z} \int _{-\infty }^{\infty } \int _{-\infty }^{\frac{z-\beta y}{\alpha }} f_\mathcal{Y _0\mathcal Y _1}(x,y)\mathrm{d}x\mathrm{d}y \nonumber \\&= \frac{1}{\alpha } \int _{-\infty }^{\infty } f_\mathcal{Y _0\mathcal Y _1}\left(\frac{z-\beta y}{\alpha },y \right)\mathrm{d}y \end{aligned}$$

(24)

Let $\mathcal Y _0$ and $\mathcal Y _1$ be independent, then their joint density $f_\mathcal{Y _0\mathcal Y _1}(x,y)=f_\mathcal{Y _0}(x)f_\mathcal{Y _1}(y)$. Now consider two random variables $\mathcal Y _0^{\prime }$ and $\mathcal Y _1^{\prime }$ with densities

$$\begin{aligned} f_\mathcal{Y _0^{\prime }}(x)=\frac{1}{\alpha }f_\mathcal{Y _0} \left( \frac{x}{\alpha } \right), \quad f_\mathcal{Y _1^{\prime }}(y)=\frac{1}{\beta }f_\mathcal{Y _1} \left( \frac{y}{\beta } \right) \end{aligned}$$

(25)

Then, (24) becomes

$$\begin{aligned} f_\mathcal{Y _s}(z)&= \frac{1}{\alpha } \int _{-\infty }^{\infty } f_\mathcal{Y _0}\left(\frac{z-\beta y}{\alpha }\right)f_\mathcal{Y _1}\left(y\right)\mathrm{d}y \nonumber \\&= \int _{-\infty }^{\infty } f_\mathcal{Y _0^{\prime }}\left(z-u\right)f_\mathcal{Y _1^{\prime }}\left(u \right)\mathrm{d}u \nonumber \\&= \left(f_\mathcal{Y _0^{\prime }} * f_\mathcal{Y _1^{\prime }}\right)(z) \end{aligned}$$

(26)

Since the characteristic function $\varphi _\mathcal{Y _s}(\omega )$ is the Fourier transform of $f_\mathcal{Y _s}(z)$, from (26), $\varphi _\mathcal{Y _0^{\prime }}(\omega ), \varphi _\mathcal{Y _1^{\prime }}(\omega )$ and $\varphi _\mathcal{Y _s}(\omega )$ are related as:

$$\begin{aligned} \varphi _\mathcal{Y _s}(\omega )=\varphi _\mathcal{Y _0^{\prime }} (\omega )\varphi _\mathcal{Y _1^{\prime }}(\omega ) \end{aligned}$$

(27)

1.2 Gaussian distribution

If $\mathcal Y _0$ and $\mathcal Y _1$ are independent and Gaussian distributed with densities $\mathcal N (\mu _0,\sigma _0^2)$ and $\mathcal N (\mu _1,\sigma _1^2)$, then from (25), $\mathcal Y _0^{\prime }$ and $\mathcal Y _1^{\prime }$ are also Gaussian distributed with densities $\mathcal N (\alpha \mu _0,\alpha ^2\sigma _0^2)$ and $\mathcal N (\beta \mu _1,\beta ^2\sigma _1^2)$, respectively. From (27), $\varphi _\mathcal{Y _s}(\omega )=\exp (j(\alpha \mu _0+\beta \mu _1) \omega -\frac{1}{2}(\alpha ^2\sigma _0^2+\beta ^2\sigma _1^2)\omega ^2)$. Taking the inverse Fourier transform, we can show that $\mathcal Y _s$ is again Gaussian distributed with density $\mathcal N (\alpha \mu _0+\beta \mu _1,\alpha ^2\sigma _0^2+\beta ^2\sigma _1^2)$.

1.3 Laplace distribution

If $\mathcal Y _0$ and $\mathcal Y _1$ are independent and Laplace distributed with densities $\mathcal L (\mu _0,\sigma _0)$ and $\mathcal L (\mu _1,\sigma _1)$, then from (25), $\mathcal Y _0^{\prime }$ and $\mathcal Y _1^{\prime }$ are also Laplace distributed with densities $\mathcal L (\alpha \mu _0,\alpha \sigma _0)$ and $\mathcal L (\beta \mu _1,\beta \sigma _1)$, respectively. Here $\mathcal Y _s$ is not Laplace distributed as in the case of Gaussian distribution. To find the density, the inverse Fourier transform can be computed as follows.

From (27), if $\alpha \sigma _0 \ne \beta \sigma _1$

$$\begin{aligned} \varphi _\mathcal{Y _s}(\omega )&= \frac{\exp (j(\alpha \mu _0+\beta \mu _1)\omega )}{(1+(\alpha \sigma _0) ^2\omega ^2)(1+(\beta \sigma _1)^2\omega ^2)} \nonumber \\&= \frac{(\alpha \sigma _0)^2}{[(\alpha \sigma _0)^2-(\beta \sigma _1)^2]} \left\{ \frac{\exp (j(\alpha \mu _0+\beta \mu _1)\omega )}{(1+(\alpha \sigma _0) ^2\omega ^2)} \right\} \nonumber \\&-\frac{(\beta \sigma _1)^2}{[(\alpha \sigma _0)^2-(\beta \sigma _1)^2]} \left\{ \frac{\exp (j(\alpha \mu _0+\beta \mu _1)\omega )}{(1+(\beta \sigma _1) \nonumber ^2\omega ^2)}\right\} \\ \end{aligned}$$

(28)

Taking the inverse Fourier transform,

$$\begin{aligned} f_\mathcal{Y _s}(x)&= \frac{1}{2[(\alpha \sigma _0)^2-(\beta \sigma _1)^2]} \nonumber \\&\times \left\{ (\alpha \sigma _0) \exp \left(-\frac{|x-(\alpha \mu _0+\beta \mu _1)|}{\alpha \sigma _0} \right) \right.\nonumber \\&-\left. (\beta \sigma _1) \exp \left(-\frac{|x-(\alpha \mu _0+\beta \mu _1)|}{\beta \sigma _1} \right) \right\} \end{aligned}$$

(29)

If $\alpha \sigma _0 = \beta \sigma _1 = \sigma$,

$$\begin{aligned} \varphi _\mathcal{Y _s}(\omega ) = \frac{\exp (j(\alpha \mu _0+\beta \mu _1)\omega )}{(1+\sigma ^2\omega ^2)^2} \end{aligned}$$

(30)

Taking the inverse Fourier transform,

$$\begin{aligned} f_\mathcal{Y _s}(x)&= \left( \frac{ \sigma + \left|x-(\alpha \mu _0+\beta \mu _1) \right| }{4\sigma ^2} \right)\nonumber \\&\times \exp \left(-\frac{|x-(\alpha \mu _0+\beta \mu _1)|}{\sigma } \right) \end{aligned}$$

(31)